Question

Question #74a3a

Answer 1

pi/2; (7pi)/6; (11pi/6)

Explanation:

`2sin t - 1 = 1/(sin t)`
`sin^2 t - sin t - 1 = 0.`
Solve this quadratic equation for sin t.
Because a + b + c = 0, use shortcut. The 2 real roots are:
sin t = 1 and `sin t = c/a = - 1/2`
Use trig table and unit circle:
a. sin t = 1 --> `t = pi/2`
b. `sin t = - 1/2` --> 2 solutions -->
`t = - pi/6`, or `t = (11pi)/6` (co-terminal), and
`t = pi - (-pi/6) = pi + pi/6 = (7pi)/6`
Answers for `(0, 2pi)`:
`pi/2; (7pi)/6; (11pi)/6`